x̄ =∑fx∑f=3
=(1×2)+(2×k)+(3×1)+(4×1)+(5×2)2+k+1+1+2=3
=2+2k+3+4+106+k=3
=19+2k6+k=3
=19+2k6+k=31
=19+2k=3(6+k)
=19+2k=18+3k
=2k-3k=18-19
=-k=-1
∴k=1